Determining Rth from a simple circuit

[x86]

Homework Statement

Homework Equations

V=IR[/B]

From my textbook:
If the circuit contains both independent and dependent sources, the open‐circuit terminals are shorted and the short‐circuit current between these terminals is determined. The ratio of the open‐circuit voltage to the short‐circuit current is the resistance Rth.

The Attempt at a Solution

I determined that Rth and Vo = 0. Now I have to find Rth. But it's not zero? What? The open circuit volage divided by the short circuit current is Rth. But we know for a fact that the open circuit voltage is 0. Rth = 0 / Ix. But this is wrong?

 

[NascentOxygen]

You determined Vo by inspection?

Have you determined Io when output is short-circuited?

You can then look at evaluating their quotient.

 

[x86]

You determined Vo by inspection?

Have you determined Io when output is short-circuited?

You can then look at doing the division.

If I'm not mistaken, I don't need to determine Io because Rth =Vth/Io = 0/Io = 0

I determined Vth/Vo by nodal analysis, it's 0 (its correct, green)

 

[gneill]

This is one of those cases where you should stick a fixed voltage source on the output and measure the resulting current that it drives into the circuit.

If you're sharp you should be able to determine almost by inspection what the resulting current will be, and thus the Thevenin resistance.

 

[x86]

This is one of those cases where you should stick a fixed voltage source on the output and measure the resulting current that it drives into the circuit.

If you're sharp you should be able to determine almost by inspection what the resulting current will be, and thus the Thevenin resistance.

Ah yes. That also was in my book, but the theory behind it wasn't explained. Where can I read about why we do this, and why it works?

 

[NascentOxygen]

If I'm not mistaken, I don't need to determine Io because Rth =Vth/Io = 0/Io = 0

Yes, you are mistaken. A non-zero value of Io is essential for this quotient to be defined.

 

[x86]

I still can't figure it out, using KCL right next to the + Vo terminal (after attaching a 1V source) I get:

Reference is the bottom node
V2 (the node in the center of the circuit, intercepted by 2 lines (looks like a +)) is -6 V
V1 = +1V

-(v2-v1)/2+v1/4-3=i

r=1/i=1.333 k ohms

This is wrong

 

[gneill]

Strange, to me it looks like 1.333 k should be correct. 4k || 2k yields 1.333k.

 

[milesyoung]

I still can't figure it out, using KCL right next to the + Vo terminal (after attaching a 1V source) I get:

Reference is the bottom node
V2 (the node in the center of the circuit, intercepted by 2 lines (looks like a +)) is -6 V
V1 = +1V

-(v2-v1)/2+v1/4-3=i

r=1/i=1.333 k ohms

This is wrong

You need to open up the independent current source and short the independent voltage source. What is your test source then connected across?

Edit:
The reasoning behind it is as follows:

Let's assume you accept Thévenin's theorem, so you can replace any two-terminal linear network as shown here:

The left circuit must then behave exactly like the right circuit, so to find R_th, we could simply turn off all the independent sources in the circuit, which leaves V_th = 0 V (a short, effectively). Your test source is then applied directly across R_th.

 

[NascentOxygen]

r=1/i=1.333 k ohms

This is wrong

If you are allowed multiple guesses, try 2kΩ

 

[x86]

If you are allowed multiple guesses, try 2kΩ

I have infinite guesses. 2k was right. How did you figure?

You need to open up the independent current source and short the independent voltage source. What is your test source then connected across?

Edit:
The reasoning behind it is as follows:

Let's assume you accept Thévenin's theorem, so you can replace any two-terminal linear network as shown here:

The left circuit must then behave exactly like the right circuit, so to find R_th, we could simply turn off all the independent sources in the circuit, which leaves V_th = 0 V (a short, effectively). Your test source is then applied directly across R_th.

In all the problems I've encountered, the book specifically says never to short a independent voltage source or open a independent current source if there are any dependent sources in the circuit. We always have had to work around them (independent is another story though)

 

[milesyoung]

In all the problems I've encountered, the book specifically says never to short a independent voltage source or open a independent current source if there are any dependent sources in the circuit. We always have had to work around them (independent is another story though)

If your book says that, then you're now taking it out of context. I can imagine it says something to that effect when it covers, for instance, source transformations, where it's very important that you don't break the dependency between independent and dependent sources.

I think if you try to work trough the reasoning I posted, you'll find that the idea behind all this is actually very, very simple. You just have to focus on the Thévenin equivalent circuit, and think about, if you had access to such a simple circuit, what tests could you come up with to determine V_th and R_th? Assume you can turn V_th on and off.

And I have to emphasize this part: Don't think in terms of the complex circuit - only think in terms of the Thévenin equivalent circuit.

 

[NascentOxygen]

Homework Statement

The mistake you've made in this exercise is that you have not followed instructions. Even were your numerical results all correct, in a written exam you could expect 0/10 because you have failed to do what was asked.

Let's review the exercise:

The instructions indicate that at the conclusion of this you would be able to determine V_o using Thévenin's theorem. But what did you do? Your very first move was to calculate V_o using a method not involving Thévenin!

EPIC FAIL!

The second instruction was that you should remove the 4kΩ resistor and calculate R_TH of the remaining circuit seen at the output terminals. This instruction could have been stated more clearly. Still, you overlooked it, so calculated for the wrong arrangement.

The required final step is to picture the Thévenin circuit driving the 4k resistor, and hence determine V_o of the circuit as drawn.

➽ There is no prohibition re "shorting out" all voltage sources and "open circuiting" all current sources here, because the circuit is drawn to trick you. The dependent source is irrelevant to V_o and R_TH. Here, V_o is wholly determined by the pair of independent sources near it. The technique of applying an external source is uncalled for.

There's a lesson in this: take careful note of instructions, and follow them to the letter!